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4 years ago

1/2 ln(x + 3) - lnx = 0 solve for x

1 answer:

6 0

1

Simplify \frac{1}{2}\imath n(x+3)21ın(x+3) to \frac{\imath n(x+3)}{2}2ın(x+3)

\frac{\imath n(x+3)}{2}-\imath nx=02ın(x+3)−ınx=0

2

Add \imath nxınx to both sides

\frac{\imath n(x+3)}{2}=\imath nx2ın(x+3)=ınx

3

Multiply both sides by 22

\imath n(x+3)=\imath nx\times 2ın(x+3)=ınx×2

4

Regroup terms

\imath n(x+3)=nx\times 2\imathın(x+3)=nx×2ı

5

Cancel \imathı on both sides

n(x+3)=nx\times 2n(x+3)=nx×2

6

Divide both sides by nn

x+3=\frac{nx\times 2}{n}x+3=nnx×2

7

Subtract 33 from both sides

x=\frac{nx\times 2}{n}-3x=nnx×2−3

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Complete the square x^2+6x-16=0

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Let's solve your equation step-by-step:
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4 years ago

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A circle has a diameter of 10 millimeters. What is the area of the circle?

professor190 [17]

Answer:

78.54 millimeters

Step-by-step explanation:

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3 years ago

The model represents an equation. What value of x makes the equation true?

timama [110]

D. x = 3

You can get this by turning these into equations

-3x + 6 = 2x - 9

Then solve.

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3 years ago

Read 2 more answers

1. The figure shows the regular triangular pyramid SABC. The base of the pyramid has an edge AB = 6 cm and the side wall has an

Musya8 [376]

Given:

• AB = 6 cm

• SM = √15 cm

Let's solve for the following:

• 1) the base elevation AM.

Given that we have a regular triangular pyramid, the length of the three bases are equal.

AB = BC = AC

BM = BC/2 = 6/2 = 3 cm

To solve for AM, which is the height of the base, apply Pythagorean Theorem:

$\begin{gathered} AM=\sqrt{AB^2-BM^2} \\ \\ AM=\sqrt{6^2-3^2} \\ \\ AM=\sqrt{36-9} \\ \\ AM=\sqrt{27} \\ \\ AM=5.2\text{ cm} \end{gathered}$

The base elevation of the pyramid is 5.2 cm.

• (2)., The elevation SO.

To find the elevation of the pyramid, apply Pythagorean Theorem:

$SO=\sqrt{SM^2-MO^2}$

Where:

SM = √15 cm

MO = AM/2 = 5.2/2 = 2.6 cm

Thus, we have:

$\begin{gathered} SO=\sqrt{(\sqrt{15})^2-2.6^2} \\ \\ SO=\sqrt{15-6.76} \\ \\ SO=2.9\text{ cm} \end{gathered}$

Length of SO = 2.9 cm

• (3). Area of the base:

To find the area of the triangular base, apply the formula:

$A=\frac{1}{2}*BC*AM$

Thus, we have:

$\begin{gathered} A=\frac{1}{2}*6^*5.2 \\ \\ A=15.6\text{ cm}^2 \end{gathered}$

The area of the base is 15.6 square cm.

• (4). Area of the side surface.

Apply the formula:

$SA=\frac{1}{2}*p*h$

Where:

p is the perimeter

h is the slant height, SM = √15 cm

Thus, we have:

$\begin{gathered} A=\frac{1}{2}*(6*3)*\sqrt{15} \\ \\ A=34.86\text{ cm}^2 \end{gathered}$

• (5). Total surface area:

To find the total surface area, apply the formula:

$TSA=base\text{ area + area of side surface}$

Where:

Area of base = 15.6 cm²

Area of side surface = 34.86 cm²

TSA = 15.6 + 34.86 = 50.46 cm²

The total surface area is 50.46 cm²

• (6). Volume:

To find the volume, apply the formula:

$V=\frac{1}{3}*area\text{ of base *height}$

Where:

Area of base = 15.6 cm²

Height, SO = 2.9 cm

Thus, we have:

$\begin{gathered} V=\frac{1}{3}*15.6*2.9 \\ \\ V=15.08\text{ cm}^3 \end{gathered}$

The volume is 15.08 cm³.

ANSWER:

• 1.) 5.2 cm

• 2.) 2.9 cm

• 3.) 15.6 cm²

• 4.) 34.86 cm²

• (5). 50.46 cm²

• 6). 15.08 cm³.

7 0

1 year ago